So taking spherical coordinates in Minkowski space as an example, √-g d 4x = r 2 sin Θ dr dΘ dφ dt and the surface element is just r 2 sin Θ dΘ dφ dt times the unit normal. For example if the surface is r = const, and r is the proper distance normal to the surface, then you get dS μ just by striking out dr from the volume element. As the Earth rotates, the fabric is distorted in the direction of.2 answers 2 votes: It is a bit more complicated since spacetime is four-dimensional. However you almost always adapt your coordinates to the surface, in which case it's quite simple. And it was predicted by Einstein over 80 years ago. It's the surface element, and indeed it can be written out in general using differential forms. Integrate by parts and you have ∫√-g V μ dS μ. There's a vector identity by which the covariant divergence of a vector can be expressed in terms of the ordinary divergence: V μ μ ≡ (1/√-g) (√-g V μ) ,μ. If you understand what the surface element is in ordinary spherical coordinates, say, then you've got it made. Vaibhavtewari, It's a lot easier than all the Hodge operator stuff! The divergence theorem is no different in four dimensions than it is in three. Hope this helps some, it's a bit rambly and not teribly rigorous. You recover the scalar volume by multiplying the dual of the three form by a unit time vector. You've probably seen volume elements represtned as 1-forms before, though it may not have been explained in these terms exactly. You can also integrate the hodges dual of the three-form, which is a one-form. Likewise, to integrate the three-form, you divide the volume element into parallel piped (like cubes, but while the faces are parallel they aren't generally at right angles), and the sides of the parallelpiped define three vectors that you feed to the three form to get a scalar, which gets summed up. The more traditional integral would be the integral of the hodges dual vector of the two form above, which gets fed the normal vector to the surface and the unit time vector instead of the two vectors that form the "sides" of a small parallelogram shaped surface element. That's the intergal of the two-form over the surface. You sum this scalar quantity over all the little trapezoids comprising the suface to and you get a scalar quantity. So you feed these two vectors that are the sides of the trapezoidal surface element into the two form that you're integrating, and you get a scalar quantity. Each small trapezoid has a natural representation as the two vectors that form the sides of the trapezoid (strictly speaking,you use the exponential map to go from the sides of the trapzoid to the vectors). So if you integrate the surface area of a 2-d surface, you break the surface up into many trapezoidal pieces. There's a natural association between a small square of area on any surface and a two-form, which describes an oriented area. The integral of the two-form described above is the integral of the hodges dual (usually described by the * operator) that doesn't use the normal vector at all. This formulation is actual the Hodges dual of the probably more familiar one which uses the normal to the surface. If we apply this to two-forms, we say that the intergal of a 2-form A over a surface is the intergal of a 3-form dA over a volume. Stokes theorem says that the intergal over a surface of a 2-form, A, is the intergal over a volume of the three-form dA, where dA is the "exterior derivative" of A. In a curved $(3 1)$ dimensional spacetime with metric components $g_\partial_\phi V^\phi \partial_zV^z.Wald, "General Relativity" talks about this some, in the apendix on Stokes theorem, in the language of differential forms.Ī n-form is an n-dimensional anti-symmetric tensor, usually described as a map from n vectors to a scalar, a map that's anti-symmetric so if you interchange any two vectors, the sign inverts.
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